Question

If tan(A) - tan(B) = x and cot(B) - cot(A) = y, then what is the value of cot(A-B)?
(a)$\dfrac{1}{x}+\dfrac{1}{y}$
(b)$\dfrac{1}{x}-\dfrac{1}{y}$
(c)$\dfrac{xy}{x+y}$
(d)$1+\dfrac{1}{xy}$

Answer 1

Hint: We will solve the given two trigonometric equations using the basic trigonometric formulae of tangent and cotangent. We will then substitute the values we get in a formula of cotangent and we will get the final answer. So, the problem requires only two concepts, namely, basic trigonometric formulae and a prior knowledge of solving equations.

Complete step-by-step answer:
Let us begin with the basic formulae of the trigonometric functions of tangent and cotangent abbreviated as tan and cot respectively.
$tan\theta =\dfrac{1}{cot\theta }..........(i)$
$\tan (A-B)=\dfrac{\tan (A)-tan(B)}{1+\tan (A)\tan (B)}..........(ii)$

Now let us use (ii) to find the value of cot(A - B).
We know that,
$\cot (A-B)=\dfrac{1}{\tan (A-B)}$
Now using the equation (ii) and substituting in the above equation we get,
$\cot (A-B)=\dfrac{1}{\dfrac{\tan (A)-tan(B)}{1+\tan (A)tan(B)}}$
Proceeding further by using the rules of reciprocation we get,
$\cot (A-B)=\dfrac{1+\tan (A)tan(B)}{\tan (A)-tan(B)}$
Now again using (i) we get,
$\cot (A-B)=\dfrac{1+\dfrac{1}{\cot (A)cot(B)}}{\dfrac{1}{\cot (A)}-\dfrac{1}{\cot (B)}}$
Taking the LCM in both the numerator and the denominator we get,
$\cot (A-B)=\dfrac{\dfrac{\cot (A)cot(B)+1}{\cot (A)cot(B)}}{\dfrac{\cot (B)-cot(A)}{\cot (A)cot(B)}}$
Simplifying by cancelling the cot(A)cot(B) in both the numerator and denominator we get,
$\cot (A-B)=\dfrac{\cot (A)cot(B)+1}{cot(B)-cot(A)}..........(iii)$
Now we have our equation ready and we already know the value of the denominator cot(B) - cot(A) as y as it is given in the question. Substituting the value y, we get,
$\cot (A-B)=\dfrac{\cot (A)cot(B)+1}{y}..........(iv)$
Now let us calculate the value of cot(A)cot(B).
In the question it is given that the value of tan(A) - tan(B) is x.
Thus, applying the formula in (i) on the both terms of the equation, we get,
tan(A)-tan(B)=x
$\Rightarrow \dfrac{1}{\cot (A)}-\dfrac{1}{\cot (B)}=x$
Taking the LCM and solving we get,
$\dfrac{\cot (B)-cot(A)}{\cot (A)cot(B)}=x$
Cross-multiplying cot(A)cot(B) we get,
$\cot (B)-cot(A)=x\cot (A)cot(B)$
But we already know the value of cot(B)-cot(A) which is y. Thus, substituting we get,
$y=x\cot (A)cot(B)$
Cross-multiplying x we get,
$\dfrac{y}{x}=\cot (A)cot(B)$
Substituting this value of cot(A)cot(B) in (iv) we get,
$\cot (A-B)=\dfrac{\dfrac{y}{x}+1}{y}$
Taking the LCM and solving we get,
$\cot (A-B)=\dfrac{y+x}{yx}$
Splitting the terms, we get,
$\cot (A-B)=\dfrac{y}{yx}+\dfrac{x}{yx}$
Cancelling the like terms, we get,
$\cot (A-B)=\dfrac{1}{x}+\dfrac{1}{y}$
Thus option(a) is the correct option.

Note: You can directly remember the formula in the equation(iii) and apply the same directly. It will save a considerable time and increase the accuracy.
Also, there is a shortcut. Since this is a multiple choice question, we can verify the options.
Let us take $A=B+\dfrac{\pi }{2}$.
Thus, we get $A-B=\dfrac{\pi }{2}$.
Now let us calculate the values of x and y.

Given, tan(A)-tan(B)=x and cot(B)-cot(A)=y
We know that $\tan \left( B+\dfrac{\pi }{2} \right)=-\tan (B)$ and $cot\left( B+\dfrac{\pi }{2} \right)=-\cot (B)$.

Putting these values, we get,
-tan(B) - tan(B) = x and cot(B) - (-cot(B)) = y
So, we get -2tan(B) = x and 2cot(B) = y
Also, cot(A-B) simplifies to $cot\left( \dfrac{\pi }{2} \right)$ which is 0.
Thus, the values of x, y and cot(A-B) are satisfied only by the option(a).